Gibbs free energy Puns

Hi everyone, I have a question that may be silly but here it goes. I understand that H or enthalpy represents the total energy of a system. I have read that that delta H, at a constant pressure represents change in heat. Also i know q, or heat flow is equal to TdeltaS. If so isn't the gibbs free energy equation just heat-heat thus 0 under a constant pressure? How do endergonic and exergonic reactions occur?

I've googled this many times and asked my professor, however, I'm still a bit confused on what standard state Gibbs Free Energy actually represents in relation to a chemical system's spontaneity and thermodynamic properties. Specifically, I'm referencing the (simplified) equation G = Enthalpy - (Temperature*Entropy), not the energy of a Galvanic cell.

I understand why the energy change of a system is the sum of heat and work, but what makes this different from Gibbs Free Energy, and why do we subtract the product of entropy and temperature from enthalpy?

For reference, I'm a chemical engineering undergrad in their first year of studies.

Looking at a derivation for the Clausius-Clapeyron equation made me a little troubled/confused.

At constant pressure and temperature the change in Gibbs free energy of a two phase system at equilibrium will be zero.

dG = Vdp - SdT + mu_1 dN_1 + mu_2 dN2 = 0

dp = dT = 0 and dN = dN_1 + dN_2 = 0 gives

(mu_1 - mu_2)dN_1 = 0 =>

mu_1 = mu_2

Okay, so far so good (I think at least).

Now at the phase boundary line in the p-T plane we have

mu1(p,T) = mu2(p,T) and moving along the boundary we also have

mu1(p + dP, T +dT) = mu2(p + dP, T +dT)

This is where my questions come in.

Because the two phases are in equilibrium (I presume) at this boundary the change in Gibbs free energy should be zero but moving along the boundary line in the p-T plane implies changes in T and p.

But we assumed that dp = dT = 0 in the beginning, right? Isn't this a problem, or does it build on two independent "facts" (phase coexistence implies that chemical potential of both phases is equal and Gibbs free energy is constant?)

I'm probably missing something but it all feels a bit circular.

My textbook lists Gibb's free energy as

Δ G = Δ H - T Δ S

where Δ H = Δ E + P Δ V

and therefore at constant temperatures, T= 0 , Δ E = 0

As a result:

Δ G = P Δ V

But in the following website: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/18%3A_Gibbs_Energy_and_Thermodynamics/18.09%3A_Gibbs_Energy_Changers_for_Non-Standard_States

It is written as:

Δ G = Δ P V

Why is there this change?

Is Gibbs free energy the maximum amount of work that a spontaneous reaction can do, or is it the maximum amount of work needed to drive a non spontaneous reaction?

Hey everyone,

I managed to find by calculation, using Gibbs free energy equation, the pKa of HCl, and I would like to do the same for the NaOH.

I started by using the equation of dissociation of NaOH in water : NaOH (+ H2O) => Na+ + OH- (+ H2O), and then calculated Delta G0 from delta H0 and delta S0.

While it worked fairly well for HCl, it gives me an absured pKa (around -7.8).

Any idea what changed between HCl and NaOH (besides the fact that one is acid and the other basic ?)

Thanks

I have been teaching chemistry at freshman college level a while, but I do find that there are gaps that arise from a combination of time out of school and specializing in one field. With my background in a weird mixture of theory and organometallic synthesis, almost all of my knowledge on thermodynamics comes down to mostly practical uses at near standard state conditions. Now that I teach more generally and my interests have focused more heavily on theory, I want to have a better understanding on how gibbs free energy relates to K.

Specifically, my actual question is: what kind of K is used in delta G=-RT ln K? By definition, standard conditions are 1 M solutions and 1 atm of gases. For a fully solution phase calculation, that means it is Kc and for a fully gas phase reaction, it is Kp. So what about in mixed phase systems? In practice, most people seem to default to Kc when doing most equilibrium calculations since it can be used in any system with a standard activity, but that doesn't work mathematically with the equation relating delta G to K. The only two solutions I have been able to surmise from the math is either the K you get is literally a mixed equilibrium constant, where all the gases are pressures and all the solutions are concentrations, or it's been way too long since I took analytical chemistry and the activity corrections fix this.

I have a text coming up tomorrow and there's a pair of things I still haven't understood about the relation ΔG=-RTln(K) at equlibrium.

1. Is K the equilibrium costant in terms of pressure ( Kp=(Pc^c*Pd^d)/(Pa^a*Pb^b) ) or in terms of concentration ( Kc=([C]^c*[D]^d)/([A]^a*[B]^b) )?
2. My textbook says that ΔG can be negative, positive, or equal to zero: what does it mean?

I hope the way I wrote the questions is clear (Pa indicates the partial pressure of gas A in a reaction like A(g) + B(g) <-> C(g) + D(g) ).

Thank you!

Im confused. Under standard conditions, gibbs free energy would be 0, since the standard cell potential is 0. But how can gibbs free energy be 0 at all concentrations, when there is a cell potential produced (can calculate using Nernst equation) and the reaction isnt in equilibrium?

I am wondering why there is a decrease in the gibbs free energy term of the volume of the crystal embryo, and why does this continue to decrease. Is there a decrease in the enthalpy term of deltaG, or an increase in the TdeltaS term?

I'm trying to understand the formula for Gibb's free energy which to my knowledge represents the amount of energy the system can use however I don't understand why you are to subtract the change in entropy of the system from the change in enthalpy of the system since a positive change in entropy would mean the system is absorbing more heat in addition to the heat it absorbs.

Also on the note of entropy, How can the change in entropy of the system be positive while the enthalpy of the system be negative?

Hi all, just trying to study for an upcoming thermo exam and I found myself stuck on a concept.

We talked about how μi = μ^(0)i(T+ + RTln(ai) in class and mentioned how μi^(0) is equivalent to Gi^(0) for an ideal gas mixture. Is that true for other chemical reactions in that μi^(0) = Gi^(0)?

I guess I'm just confused as to the practical purpose of having that μ term in equations since we did derive that dG = RTlnK with that μ definition. Sorry if this question is confusing!

I already read my book, searched and tried to understand from previous eli5 on this, but I stilldon't get it. enthalpy and temperature seem to be the same or very similar things, so why are they both in the equation but written as two terms? I know the baseline for temperature in kelvin is 0, as in absolute zero. How is enthalpy measured? I thought it was heat of formation but now I think thats ΔG. And how is entropy measured and what are its reference values (like how temperature has absolute zero). Why is the units joules per kelvin, what kind of sense does that make? ΔG determines how likely the reaction is to take place, but I thought that was ΔS. Basically all the things seem to blend into each other and I can't figure out what's the point of any of it

ΔG= ΔH – TΔS

Thanks

So I saw these derivations in a textbook:

>dU = ∂w + ∂q
>
>Substitute dS >= ∂q/T and ∂w = - PdV to obtain:
>
>dU <= TdS - PdV
>
>At constant P and V:
>
>d(U + PdV - TdS) <= 0
>
>Define G = U + PV - TS. The above equation becomes:
>
>dG <= 0 (constant T and P)

The authors only substitute ∂w = - PdV, although I learned that ∂W equals the sum of all the work done by the system. So does that mean if the system does more than just P-V work, we can't use dG to determine the spontaneity of the system processes anymore?Thank you very much for reading.

Derivation of gibbs free energy according to my textbook

I understand everything except how it got to equation 22.10 from the previous step. I do understand that T and P are constant....but to my barely-remember-calculus brain can't understand why the dV becomes V.....

I suspect there is some rule about manipulating differential equations that I'm missing, if anyone knows a good textbook dealing with this please name drop it.

Can a rock rolling down a hill be described as 'spontaneous' in Gibbs term?

To streth things even more, can humanity's progress toward secularim be described as an endothermic reaction?

what is change in g for the autoionization of water at 25 celcius? gives R= 8.314462

For the Gibbs Free Energy equation, my textbook says that the change in entropy multiplied by the temperature represents the total amount of energy that is absorbed by a system when its entropy increases reversibly.

Can someone explain this to me? I have a hard time understanding what this means. Thanks!

Hello, complete amateur here. When looking at one form of the equation deriving the change in Gibbs free energy for a given chemical reaction, the terms change in Enthalpy, and the negative product of Temperature and change in Entropy:

*(delta)G = (delta)H - T(in Kelvin)*(delta)*S

This seems to double count how Temperature will influence the spontaneity of the reaction if Temperature is already a consideration when deriving Entropy, S.

Any guidance would be appreciated :)

So Gibbs‘ free energy is the maximum reversible work that can be produced in a thermodynamic system at constant P and T, and according to the formula, higher T means lower ΔG. Then why is Superheated steam used as means to produce work? Or am I looking at it wrong?

I'm currently studying for the MCAT and I just got to Gibbs's free energy equation. This was a concept I never fully grasped when I was taking genchem 2. I am trying to understand the equation conceptually and it just won't click for me. I got through my tests in chemistry by just memorizing what positive and negative (delta)G meant. However, I am hoping someone can really dumb this down to a level I can understand. My review book defines Gibbs free energy as, "the maximum amount of work that can be performed by a system consisting of one or more reversible chemical reactants."

I understand that a positive delta G means a reaction is non-spontaneous and therefore not energetically favored, or is this even right? Which makes sense in that it would be an endothermic rxn. So in my head when I think about an endothermic rxn I think about bonds being formed which would decrease entropy???

The reverse would be true for a negative delta G, making the reaction spontaneous and energetically favorable. Which makes sense if I look at a rxn coordinate or think about a campfire, it would take an initial spark to get over an energy barrier and the resulting enthalpy of bonds breaking would keep the fire going. I guess this would increase the entropy or "disorder"???

So, ATP hydrolysis is exergonic reaction(spontaneous) because the ATP molecule is broken down in order provide energy correct? So energy is release!!! But in term of enthalpy which talked about bond formation. ATP hydrolysis is breaking down bonds... so it would be an endothermic process? And vice versus for ATP regeneration. Is my reasoning right. Or I’m not supposed to think of Bond Enthalpy when thinking about ATP hydrolysis?

So basically i found a website that has the derivation but maybe I'm just bad at maths but i don't understand this step

i understand why we integrate but I don't understand how to get from the third to the fourth line

Hello everyone,

I'd like to ask a question about entropy and gibbs free energy, i really can't get my head around it! Say we have an exothermic reaction and the entropy change is positive; then dG = dH - Tds is negative... But besides dH; there's a positive change in entropy so - TdS is negative. That would mean dG decreases even further; which means more free energy is available. Does this mean the entropy term can be used to get mechanical work out of it?

for example, an explosion, lets say in a diesel engine -> let us ignore the enthalpy change. Let us focus on the entropy term (-Tds); the entropy increases because the molecules can disperse more freely since they've been converted to gas. Now, does this mean the entropy alone can be used to do work?

If not, what does the entropy mean for an exothermic reaction (or if you could give an example from an endothermic reaction too, i'd be pleased!)? it isn't the energy used from the enthalpy (heat) to disperse because otherwise the gibbs free energy would increase again. So what does it mean exactly, then?

Thanks in advance!